1. **State the problem:** We have 60 class sizes from statistics classes and need to construct a Frequency Distribution Table (FDT) and calculate mean, median, mode, variance, standard deviation, and coefficient of variation. 2. **Organize data:** The data set is: $$\{31,18,46,57,32,36,30,61,58,20,48,22,52,26,43,25,40,58,29,58,27,44,15,40,33,28,52,33,47,38,33,26,43,39,54,51,57,38,56,49,21,22,19,63,54,42,35,31,42,43,49,55,38,17,34,35,30,29,14,18\}$$ 3. **Construct Frequency Distribution Table (FDT):** - Find range: max = 63, min = 14, range = 63 - 14 = 49 - Choose class width (approximate): $\frac{49}{10} = 4.9 \approx 5$ - Classes (intervals): 14-18, 19-23, 24-28, 29-33, 34-38, 39-43, 44-48, 49-53, 54-58, 59-63 | Class Interval | Frequency | |---------------|-----------| | 14 - 18 | 5 | | 19 - 23 | 5 | | 24 - 28 | 6 | | 29 - 33 | 7 | | 34 - 38 | 7 | | 39 - 43 | 6 | | 44 - 48 | 4 | | 49 - 53 | 5 | | 54 - 58 | 7 | | 59 - 63 | 3 | 4. **Calculate midpoints ($x_i$) for each class:** $$x_i = \frac{\text{lower limit} + \text{upper limit}}{2}$$ | Class Interval | Midpoint $x_i$ | Frequency $f_i$ | $f_i x_i$ | $f_i x_i^2$ | |---------------|----------------|-----------------|-----------|-------------| | 14 - 18 | 16 | 5 | 80 | 1280 | | 19 - 23 | 21 | 5 | 105 | 2205 | | 24 - 28 | 26 | 6 | 156 | 4056 | | 29 - 33 | 31 | 7 | 217 | 6727 | | 34 - 38 | 36 | 7 | 252 | 9072 | | 39 - 43 | 41 | 6 | 246 | 10086 | | 44 - 48 | 46 | 4 | 184 | 8464 | | 49 - 53 | 51 | 5 | 255 | 13005 | | 54 - 58 | 56 | 7 | 392 | 21952 | | 59 - 63 | 61 | 3 | 183 | 11163 | 5. **Sum frequencies and products:** $$\sum f_i = 60$$ $$\sum f_i x_i = 2086$$ $$\sum f_i x_i^2 = 88605$$ 6. **Calculate mean ($\bar{x}$):** $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2086}{60} \approx 34.77$$ 7. **Calculate median:** - Median class is where cumulative frequency reaches $\frac{60}{2} = 30$. - Cumulative frequencies: 5, 10, 16, 23, 30, ... median class is 34-38. - Median formula: $$\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f_m}\right) \times w$$ Where: - $L=33.5$ (lower boundary of median class) - $N=60$ - $F=23$ (cumulative frequency before median class) - $f_m=7$ (frequency of median class) - $w=5$ (class width) Calculate: $$\text{Median} = 33.5 + \left(\frac{30 - 23}{7}\right) \times 5 = 33.5 + \frac{7}{7} \times 5 = 33.5 + 5 = 38.5$$ 8. **Calculate mode:** - Mode class is the class with highest frequency, which is 7 (multiple classes: 29-33, 34-38, 54-58). - Choose first mode class 29-33. - Mode formula: $$\text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times w$$ Where: - $L=28.5$ (lower boundary of mode class) - $f_1=7$ (frequency of mode class) - $f_0=6$ (frequency before mode class) - $f_2=7$ (frequency after mode class) - $w=5$ Calculate: $$\text{Mode} = 28.5 + \frac{7 - 6}{2 \times 7 - 6 - 7} \times 5 = 28.5 + \frac{1}{14 - 13} \times 5 = 28.5 + 5 = 33.5$$ 9. **Calculate variance ($\sigma^2$):** $$\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \bar{x}^2 = \frac{88605}{60} - (34.77)^2 = 1476.75 - 1208.43 = 268.32$$ 10. **Calculate standard deviation ($\sigma$):** $$\sigma = \sqrt{268.32} \approx 16.38$$ 11. **Calculate coefficient of variation (CV):** $$\text{CV} = \frac{\sigma}{\bar{x}} \times 100 = \frac{16.38}{34.77} \times 100 \approx 47.11\%$$ **Final answers:** - Mean $\approx 34.77$ - Median $= 38.5$ - Mode $= 33.5$ - Variance $\approx 268.32$ - Standard deviation $\approx 16.38$ - Coefficient of variation $\approx 47.11\%$